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Indian Institute of Technology, Bombay Chemical Engineering

CL603: Optimization Quiz 2, Spring 2009

Follow   ALL   instructions carefully: This is a   CLOSED BOOK  exam. Calculators are permitted. Make reasonable

assumptions and  CLEARLY  indicate them in your answer book. Where possible, box your final answers. Start each

problem on a new sheet.

1. A thin-walled cylindrical pressure vessel with spherical ends is to be designed to minimize the total volume

of the material used in its manufacture. The vessel must contain at least 25 m     of gas at a pressure,     , of 

0.35 MPa. The hoop stress,  

 

  , in the cylinder walls must not exceed 200 MPa ( 

 

       ). A schematic

of the vessel is provided below.

(a) Graphically identify the feasible region on an 

  vs. 

  plot. [2]

(b) Using the Kuhn-Tucker conditions (i.e first order conditions), determine the values of radius 

  and

wall thickness 

  which minimize the volume of the material subject to the constraints. [6]

(c) Calculate the sensitivity of the optimal solution to small changes in the right hand sides of the

inequality constraints. [2]

Soln:

1. Let     

  . We want to minimize 

 

       

 

   

  where   

  m, subject to 

 

         

  MPa

(      

  Mpa) and 

 

       

 

   

 

       

   . Then

             

 

   

 

 

     

   

 

         

 

      

 

 

 

 

   

 

     

 

 

                       

. 

 

           

  . The two constraints intersect at       

     m.

2. KKT is     

x 

   

 

 

  T 

g 

 x 

 

  0

 

 

 

Tg    x 

   0

 

 

   

 

       

 

 

   

 

       

 

 

 

   

 

 

   

 

     

 

 

   

 

   

 

 

     

 

       

 

 

 

 

 

 

 

 

 

 

 

 

   

 

 

 

     

 

   

 

 

 

 

   

   

   

Which results in 4 cases 

 

 

 

 

 

 

 

         

     

       

   

A:  

 

   

 

                   and                which implies      or            and      or

      and hence no feasible solution exists.

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B:  

 

   and hence                

 

               and                and        

 

       

 

     .

The last eq. gives     

  m while the second gives   

  or         

  and hence no feasible

solution again.

C: 

 

                   

 

     

  and                 

 

   

 

 

  and     

  . Taking ratio of 

first two eqs,                                 

  which gives         

  which is not feasible.

D:  

 

   

 

   .  

 

   gives        m.  

 

   gives          

   

  m.

               

 

       

 

             

                 

 

   

 

 

Second eq gives 

 

 

               

 

 

       

   

   

  and first gives

 

 

 

               

 

   

             

       

   

   

and hence we have a global minimum.

3.     

  x 

 

  d  

  at  d  

  is     

 

 

  . and hence the sensitivities are calculated already from above. The solutionis more sensitive to changes in the second constraint (volume) rather than the first (hoop stress).